# Electrostatic Potential and Field for an Infinite Conducting Plane with Charges on Both Sides

Suppose that we have an infinite conducting plate that occupies the $$x-z$$ plane. We put 3 point charges $$q_1 = Q,\quad q_2 = Q$$ and $$q_3=2Q$$ in the points $$(0,a,b),\quad (0,a,-b)$$ and $$(0,-a,0)$$ respectively. So in the $$y-z$$ plane the system of 3 charges and the plate will look like the image below.

We want to find the electric potential and field everywhere. The first question is, can we solve this problem using the method of images? If not, the most general solution to the boundary-value electrostatic problem can be given using the Green's functions. How can we find the Green's function for this problem?

Assuming that we can use the method of images, another question arises. We know that the image charges will be $$q'_1=-Q,\quad q'_2=-Q$$ and $$q'_3=-2Q$$ located at $$(0,-a,b),\quad (0,-a,-b)$$ and $$(0,a,0)$$ respectively in order to keep the potential at zero on the $$x-z$$ plane. Now, if we want to calculate the electric field at $$(0,a,0)$$, should we disregard the image charge $$q'_3$$ which is located at that point? Similarly, if we want to calculate the electric field at $$(0,-a,b)$$ or $$(0,-a,-b)$$, should we forget about the image charges $$q'_1$$ and $$q'_2$$ in order to carry out our calculations?

• Is the plate grounded, or just conducting? Commented Sep 7, 2021 at 15:18
• @MichaelSeifert being grounded means it has the same potential as infinite, thus, being the plane infinite itself, it must be grounded Commented Sep 7, 2021 at 15:52
• @hellofriends: That makes sense; I just wasn't sure that I wasn't missing some important detail. Commented Sep 7, 2021 at 15:53

Since the plane is conducting and it "touches" infinity, we can assume that its potential is the same as the potential at infinity; in other words, it is grounded. And if the plane is grounded, then none of the charges create a potential or an electric field on the opposite side of the plane. This means that we can solve for the two sides of the plane separately and add the solutions together.

The method is as you described in your last paragraph. The potential can be viewed as the superposition of the potential from two simpler problems:

• $$\phi_+(x,y,z)$$ will be the potential generated by the charges at $$(0, a, \pm b)$$ with the boundary condition $$\phi_+(x,0,z) = 0$$, ignoring the charge with $$y<0$$ completely. The solution to this problem in the region $$y > 0$$ can be solved with the method of images, using two image charges of $$-Q$$ at $$(0,-a,\pm b)$$; and for $$y < 0$$ we will have $$\phi_+ = 0$$.

• Similarly, $$\phi_-(x,y,z)$$ will be the potential generated by the charge at $$(0, -a, 0)$$ with the boundary condition $$\phi_+(x,0,z) = 0$$, ignoring the charges at $$(0,a,\pm b)$$. We can solve for $$\phi_-$$ in the region $$y<0$$ using an image charge of $$-2Q$$ at $$(0,a,0)$$, and for $$y>0$$ we will have $$\phi_- = 0$$.

The total potential will then be $$\phi(x,y,z) = \phi_+(x,y,z) + \phi_-(x,y,z)$$. The superposition of these two solutions still satisfies the boundary condition $$\phi(x,0,z) = 0$$, and it's not too hard to see that $$\nabla^2 \phi = \rho/\epsilon_0$$, where $$\rho$$ is the overall charge distribution. (Effectively, in the above method we split $$\rho$$ into the superposition of two charge configurations $$\rho_+$$ & $$\rho_-$$, with $$\nabla^2 \phi_+ = \rho_+/\epsilon_0$$ and $$\nabla^2 \phi_- = \rho_-/\epsilon_0$$.)

• I did not understand why in the case of $y>0$ you not include the charge $2Q$ and same with the case of $y<0$. the boundary conditions must be satisfied for method of images Commented Sep 8, 2021 at 10:50
• @Sagigever: I've done a major rewrite that explains why the boundary conditions are still satisfied using this method. Commented Sep 8, 2021 at 11:46

My answer is no. You cannot use image charge on this problem.

The principle that make the image-charge method work is the unique theorem. Under this principle, a solution (a unique one) must follows two conditions:

1. It must be a solution of the Poisson's equation in this region: $$\nabla^2 \psi(\vec r) = -\rho(\vec r).$$
2. it satisfies the boundary (which encloses this region) condition.

For your image charge scheme, the charge densities $$\rho(\vec r)$$ in both side are different from the original charge distribution. Therefore, your solution won't be the correct solution for both side.

• Thanks for the answer. How can we construct the Green's function for this problem in order to express the electrostatic potential? Commented Sep 7, 2021 at 14:31
• You can find Green's function $G(\vec r, \vec r')$ for $\vec r$ and $\vec r'$ loacte in a same side using image charge method. But for Green's function in the other side, you cannot use image charge method. You have to solve Laplace equation and match Neumann boundary condition.
– ytlu
Commented Sep 7, 2021 at 14:44
• Note that if the plane is grounded, then $G(\vec{r},\vec{r}') = 0$ when $\vec{r}$ and $\vec{r}'$ are on opposite sides of the plane. Proof: if there is no charge on one side of the plane, then $\phi = 0$ on that side of the plane regardless of what the charge configuration on the other side of the plane is. This follows because $\phi = 0$ is the solution to Laplace's equation in a half-space with $\phi = 0$ on the boundary and $\phi =\to 0$ at infinity. Commented Sep 7, 2021 at 15:31

Interesting, let's think it through with an example.

As ytlu mentioned, initial conditions must be met by the offered solution (boundary conditions). if you know the classic example of the charged particle and the conductive sheet, the solution that you will is only true for one region of space. Here you have charges on both sides, so this won't work out.

If not, how can we solve it?

Let's assume the conductive sheet is grounded. so you choose a random point on it, and you know for a fact that the potential is going to be zero. What are the sources of electric fields present in our setup? The three charged particles and the charge on the plane.

So you can easily calculate the potential caused by the three particles. let call that

$$\phi_1$$

Now, how does the charge distribution affect the potential? A potential caused by the charges at a random point (x,z) can be calculated as

$$d\phi_2 = \frac{\sigma(r,\theta)rdrd\theta}{4\pi\epsilon_0r}$$

where:

$$r = \sqrt{x^2+z^2}$$

If you can't see it, the differential terms in the fraction are just denoting the area of a very small square on the sheet. If you are familiar with calculus, you will see that

$$\phi_2 = \frac{1}{4\pi\epsilon_0} \int\int \sigma(r,\theta)\, drd\theta$$

From here, you can write down:

$$\phi_1 + \phi_2 = 0$$

Ending up with the expression:

$$-4\pi\epsilon_0\phi_1(r,\theta) = \int\int\sigma(r,\theta)\,drd\theta$$

I think going further than this for me would not be about the focus of your problem. Hope this helps. From here on, you have the potential caused by the charges and can solve for the charge density, then using that to come up with expressions of potential in your space.

Addition: if the plate is not grounded, you have another equation about the charges and how the overall sum of the charges should be 0...